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Suppose that people arrive at a bus stop in accordance with a Poisson process with rate \(\lambda\). The bus departs at time \(t\). Let \(X\) denote the total amount of waiting time of all those who get on the bus at time \(t\). We want to determine \(\operatorname{Var}(X)\). Let \(N(t)\) denote the number of arrivals by time \(t\). (a) What is \(E[X \mid N(t)] ?\) (b) Argue that \(\operatorname{Var}[X \mid N(t)]=N(t) t^{2} / 12\) (c) What is \(\operatorname{Var}(X) ?\)

Short Answer

Expert verified
The variance of the total waiting time, Var(X), is given by \(\frac{\lambda t^{3}}{3}\).

Step by step solution

01

(a) Finding E[X | N(t)]

Let's find the expected value of X given N(t), the number of arrivals by time t. The waiting time for each person is uniformly distributed between 0 and t. Therefore, we have: E[X | N(t)] = N(t) * E[waiting time for each person] = N(t) * (t / 2), since the expected value of a uniform distribution over the range [0, t] is t/2.
02

(b) Finding Var[X | N(t)]

Now, let's find the conditional variance Var[X | N(t)]. Since the waiting times are independent and uniformly distributed, we consider the variance of the waiting time for each person and sum it up N(t) times: Var[X | N(t)] = N(t) * Var[waiting time for each person] = N(t) * (t^2 / 12), since the variance of a uniform distribution over the range [0, t] is t^2/12.
03

(c) Finding Var(X)

Now, we can find the variance of X by using the conditional variance that we found in step (b) and the conditional expectation that we found in step (a) by applying the law of total variance: Var(X) = E[Var[X | N(t)]] + Var(E[X | N(t)]) First, we find E[Var[X | N(t)]]: E[Var[X | N(t)]] = E[N(t) * (t^2 / 12)] = E[N(t)] * (t^2 / 12) = (λt) * (t^2 / 12), since E[N(t)] = λt as N(t) follows a Poisson distribution with rate λ. Next, we find Var(E[X | N(t)]): Var(E[X | N(t)]) = Var(N(t) * (t / 2)) = (t / 2)^2 * Var(N(t)) = (t / 2)^2 * (λt), since Var(N(t)) = λt as N(t) follows a Poisson distribution with rate λ. Finally, substituting the results from above: Var(X) = (λt) * (t^2 / 12) + (t / 2)^2 * (λt) = (λt^3/12) + (λt^3/4) = (λt^3/3). Therefore, the variance of the total waiting time, Var(X), is given by \(\frac{\lambda t^{3}}{3}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conditional Expectation
The concept of conditional expectation is pivotal in understanding complex random processes. It describes the average outcome of a random variable given that certain conditions are met. In the case of the bus stop problem, the conditional expectation, denoted as E[X | N(t)], represents the expected total waiting time of all people at the bus stop given the number of arrivals by a certain time.
Taking into consideration that the waiting time for each individual is uniformly distributed between 0 to t, each person's average waiting time is simply the midpoint of that interval, which is t/2. Therefore, if N(t) people arrive by time t, the total expected waiting time will be N(t) multiplied by t/2. This method of calculation helps ensure that predictions about the system's behavior are accurate and reflective of all possible outcomes.
Conditional Variance
While conditional expectation gives us an average, conditional variance measures the spread or variability around that average, under certain conditions. In our Poisson process example, once we know the number of arrivals N(t), we seek to determine how much the total waiting time X can fluctuate around its expected value.
For independent and uniform waiting times, the conditional variance is calculated by adding the variances of individual waiting times N(t) times. Since the variance for a uniform distribution between 0 and t (t^2/12) reflects the spread of an individual's waiting time, the total spread or conditional variance for N(t) arrivals is N(t) times t^2/12. This offers a quantifiable way to gauge uncertainty in the system, an essential consideration for predicting and managing waiting times at the bus stop.
Uniform Distribution
In statistics, a uniform distribution is a type of probability distribution in which all outcomes are equally likely. The waiting times at the bus stop are an example of a continuous uniform distribution, as they can take any value between 0 and t with equal likelihood. Two key characteristics of this distribution used in our problem are the mean (t/2) and the variance (t^2/12).

Properties of Uniform Distribution

  • Mean: The average value is the midpoint of the range, making calculations straightforward.
  • Variance: Given by the formula (end point - start point)^2/12 which quantifies the extent to which individual data points spread out from the mean.
These characteristics are instrumental in establishing the building blocks for more complex calculations in our problem, like finding the total expected waiting time and its variability.
Law of Total Variance
The law of total variance breaks down the total variance of a random variable into two parts: the expected value of the conditional variances and the variance of the conditional means. This law is represented by the equation Var(X) = E[Var[X | Y]] + Var(E[X | Y]), where X is the random variable of interest and Y is another random variable upon which X is conditioned.
In the bus stop scenario, we apply this law by first calculating the expected conditional variance, which incorporates the variability of individual wait times, and then computing the variance of the conditional expectations, representing the variability in the number of arrivals. These calculations allow us to integrate both sources of uncertainty to find the total variance of the waiting time, offering a complete picture of the variability in the system.
Random Arrival
The notion of random arrival is central to models like the Poisson process, which describes events occurring randomly over time. In our exercise, the arrivals at the bus stop by time t is a textbook case of a Poisson process. By designating arrivals as random, we acknowledge that they are independent of each other and the probability of an arrival in any given interval is the same.
Understanding the randomness of arrivals is essential for using statistical techniques like the Poisson distribution to predict how many individuals will arrive in a set period and informs the use of conditional expectations and variances. It ensures we have an adaptable and robust methodology for quantifying and managing the intrinsic randomness encountered in real-world scenarios, such as bus stop arrivals.

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Most popular questions from this chapter

In Example \(5.3\) if server \(i\) serves at an exponential rate \(\lambda_{i}, i=1,2\), show that \(P\\{\) Smith is not last \(\\}=\left(\frac{\lambda_{1}}{\lambda_{1}+\lambda_{2}}\right)^{2}+\left(\frac{\lambda_{2}}{\lambda_{1}+\lambda_{2}}\right)^{2}\)

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Suppose that electrical shocks having random amplitudes occur at times distributed according to a Poisson process \(\\{N(t), t \geqslant 0\\}\) with rate \(\lambda .\) Suppose that the amplitudes of the successive shocks are independent both of other amplitudes and of the arrival times of shocks, and also that the amplitudes have distribution \(F\) with mean \(\mu\). Suppose also that the amplitude of a shock decreases with time at an exponential rate \(\alpha\), meaning that an initial amplitude \(A\) will have value \(A e^{-\alpha x}\) after an additional time \(x\) has elapsed. Let \(A(t)\) denote the sum of all amplitudes at time \(t\). That is, $$ A(t)=\sum_{i=1}^{N(t)} A_{i} e^{-\alpha\left(t-S_{i}\right)} $$ where \(A_{i}\) and \(S_{i}\) are the initial amplitude and the arrival time of shock \(i\). (a) Find \(E[A(t)]\) by conditioning on \(N(t)\). (b) Without any computations, explain why \(A(t)\) has the same distribution as does \(D(t)\) of Example \(5.21\).

Consider a two-server system in which a customer is served first by server 1, then by server 2, and then departs. The service times at server \(i\) are exponential random variables with rates \(\mu_{i}, i=1,2 .\) When you arrive, you find server 1 free and two customers at server 2 -customer \(\mathrm{A}\) in service and customer \(\mathrm{B}\) waiting in line. (a) Find \(P_{A}\), the probability that \(A\) is still in service when you move over to server 2 . (b) Find \(P_{B}\), the probability that \(B\) is still in the system when you move over to server 2 . (c) Find \(E[T]\), where \(T\) is the time that you spend in the system. Hint: Write $$ T=S_{1}+S_{2}+W_{A}+W_{B} $$ where \(S_{i}\) is your service time at server \(i, W_{A}\) is the amount of time you wait in queue while \(A\) is being served, and \(W_{B}\) is the amount of time you wait in queue while \(B\) is being served.

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